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Chapter 19: Problem 6

The magnetic field of Earth at a certain location is directed vertically downward and has a magnitude of \(50.0 \mu \mathrm{T}\). A proton is moving horizontally toward the west in this field with a speed of \(6.20 \times 10^{6} \mathrm{~m} / \mathrm{s}\). What are the direction and magnitude of the magnetic force the field exerts on the proton?

Short Answer

Expert verified
The magnitude of the magnetic force the field exerts on the proton is \(5.0 x 10^{-14} N\) and the direction of this force is South.

Step by step solution

01

Determine the Given Values

The charge of the proton (q) is \(1.602 x 10^{-19} C\). As stated, the velocity (v) of the proton is \(6.20 x 10^6 m/s\) and the magnetic field strength (B) is \(50.0 x 10^{-6} T\). Since the proton is moving horizontally and the magnetic field is vertical, the angle between v and B (胃) is \(90掳\).
02

Apply the Magnetic Force Formula

Now, using the equation for magnetic force \(F = qvBsin胃\), plug in the given values. Here, \(sin 90掳 = 1\), so the equation simplifies to \(F = qvB\).
03

Calculate the Magnitude of the Force

Substituting the given values into the equation results in \(F = (1.602 x 10^{-19} C) * (6.20 x 10^6 m/s) * (50.0 x 10^{-6} T)\). Solving this yields \(F = 5.0 x 10^{-14}\) Newtons.
04

Determine the Direction of the Force

The right-hand rule states that if you extend your right hand with your finger in the direction of positive charge movement (velocity) and your palm facing in the direction of the magnetic field, your thumb points in the direction of the force. In this case, if you extend your right fingers to the West and bend your fingers downwards, your thumb points to the South. So the magnetic force is directed to the South.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Magnetic Fields
Magnetic fields are invisible forces that emanate from magnets or moving charged particles. They are represented by magnetic field lines. These lines depict both the strength and direction of the field. The strength of the magnetic field is measured in Tesla (T). One Tesla is a quite strong field; Earth's magnetic fields are much smaller, typically around the microtesla range. In the example exercise, the magnetic field magnitude was given as 50.0 microtesla (or 50.0 x 10^{-6} T), which is quite common near Earth's surface.
Magnetic fields exert a force on moving charged particles. This force depends on the magnitude of the charge, the velocity of the particle, the strength of the magnetic field, and the angle between the direction of the velocity and magnetic field. When these forces act, they affect the path of the particle without changing its speed.
Right-Hand Rule
The right-hand rule is a simple yet essential tool to determine the direction of the magnetic force acting on a charged particle. It is particularly influential in physics and engineering-related disciplines as it offers a visual way of understanding magnetic interactions.
To apply the right-hand rule, follow these steps:
  • Point your fingers in the direction of the velocity of the positive charge (for protons) or opposite for negative charges (like electrons).
  • Align your palm to face the direction of the magnetic field lines.
  • Your thumb will then point in the direction of the magnetic force exerted on the particle.
In the given exercise, the velocity of the proton is directed horizontally towards the West, and the magnetic field is directed vertically downward. Using the right-hand rule, the force exerted on the proton would be directed to the South. This visualization helps in predicting the behavior of charged particles in magnetic fields.
Proton Motion in Magnetic Fields
When a proton, a positively charged particle, moves through a magnetic field, it experiences a force that can change its direction of motion. According to the scenario right out of the original exercise, the force does not depend on the particle's path being linear. Instead, it makes the proton follow a curved trajectory.
This is because the magnetic force acts perpendicular to both the velocity of the proton and the magnetic field. As a result, the work done by the magnetic field on the proton is zero, meaning it doesn鈥檛 speed up or take away the particle's kinetic energy. It only influences the proton's direction.
The equation to calculate this magnetic force is given by: \[ F = qvB\sin\theta \]Where:
  • \( F \) is the force,
  • \( q \) is the charge of the proton (\( 1.602 \times 10^{-19} \) C),
  • \( v \) is the particle's velocity,
  • \( B \) is the magnetic field strength,
  • \( \theta \) is the angle between the velocity and the magnetic field.
In the exercise, this led to a result of 5.0 x 10^{-14} Newtons being the magnitude of force acting southwards, bending the path of the proton within the magnetic field.

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Most popular questions from this chapter

An eight-turn coil encloses an elliptical area having a major axis of \(40.0 \mathrm{~cm}\) and a minor axis of \(30.0 \mathrm{~cm}\) (Fig. P19.27). The coil lies in the plane of the page and has a \(6.00-\mathrm{A}\) current flowing clockwise around it. If the coil is in a uniform magnetic field of \(2.00 \times 10^{-4} \mathrm{~T}\) directed toward the left of the page, what is the magnitude of the torque on the coil? (Hint: The area of an ellipse is \(A=\pi a b\), where \(a\) and \(b\) are, respectively, the semimajor and semiminor axes of the ellipse.)

In 1962 measurements of the magnetic field of a large tornado were made at the Geophysical Observatory in Tulsa, Oklahoma. If the magnitude of the tornado's field was \(B=1.50 \times 10^{-8} \mathrm{~T}\) pointing north when the tornado was \(9.00 \mathrm{~km}\) east of the observatory, what current was carried up or down the funnel of the tornado? Model the vortex as a long, straight wire carrying a current.

A wire with a mass of \(1.00 \mathrm{~g} / \mathrm{cm}\) is placed on a horizontal surface with a coefficient of friction of \(0.200 .\) The wire carries a current of \(1.50 \mathrm{~A}\) eastward and moves horizontally to the north. What are the magnitude and the direction of the smalles vertical magnetic field that enables the wire to move in this fashion?

A cosmic-ray proton in interstellar space has an energy of \(10.0 \mathrm{MeV}\) and executes a circular orbit having a radius equal to that of Mercury's orbit around the Sun, which is \(5.80 \times 10^{10} \mathrm{~m}\). What is the magnetic field in that region of space?

An electron is moving at a speed of \(1.0 \times 10^{4} \mathrm{~m} / \mathrm{s}\) in a circular path of radius \(2.0 \mathrm{~cm}\) inside a solenoid. The magnetic field of the solenoid is perpendicular to the plane of the electron's path. Find (a) the strength of the magnetic field inside the solenoid and (b) the current in the solenoid if it has 25 turns per centimeter.
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